The wavelength of the note emitted by a tuning fork of frequency 100 Hz is 3.32 m. if the density of air at STP is 1.29 `kg//m^(3)`, the `gamma` for air is

Options

A. 1.57
B. 2.3
C. 1.41
D. 1.732

Answer

To solve the problem, we need to find the value of gamma (γ) for air given the frequency of the tuning fork, the wavelength of the note emitted, and the density of air at standard temperature and pressure (STP).

### Step-by-Step Solution:

1. **Identify the given values:**
- Frequency (n) = 100 Hz
- Wavelength (λ) = 3.32 m
- Density of air (ρ) = 1.29 kg/m³
- Pressure at STP (P) = 1.013 × 10^5 N/m²

2. **Calculate the velocity (V) of sound:**
The formula for the velocity of sound is given by:
\[ V = n \cdot \lambda \]
Substituting the values:
\[ V = 100 \, \text{Hz} \times 3.32 \, \text{m} = 332 \, \text{m/s} \]

3. **Use the formula for the velocity of sound in terms of gamma:**
The velocity of sound can also be expressed as:
\[ V = \sqrt{\frac{\gamma P}{\rho}} \]
Squaring both sides gives:
\[ V^2 = \frac{\gamma P}{\rho} \]

4. **Rearranging to find gamma (γ):**
From the equation above, we can express gamma as:
\[ \gamma = \frac{V^2 \cdot \rho}{P} \]

5. **Substituting the known values:**
- \( V^2 = (332 \, \text{m/s})^2 = 110224 \, \text{m}^2/\text{s}^2 \)
- Substitute \( \rho = 1.29 \, \text{kg/m}^3 \) and \( P = 1.013 \times 10^5 \, \text{N/m}^2 \):
\[ \gamma = \frac{110224 \cdot 1.29}{1.013 \times 10^5} \]

6. **Calculate gamma:**
\[ \gamma = \frac{142.5}{101300} \approx 1.40 \]

### Final Answer:
The value of gamma (γ) for air is approximately **1.40**.